Question

Giải hộ mình câu f với các bạn.

Answer 1

Đặt \(\dfrac{1}{2x+3y}\)=u; \(\dfrac{1}{2x-3y}\)=v

=> hệ pt <=> \(\left\{{}\begin{matrix}2u-3v=-\dfrac{3}{5}\\9u+8v=\dfrac{67}{15}\end{matrix}\right.\)

=> giải hpt =>\(\left\{{}\begin{matrix}u=\dfrac{1}{5}\Rightarrow\dfrac{1}{2x+3y}=\dfrac{1}{5}\Rightarrow2x+3y=5\\v=\dfrac{1}{3}\Rightarrow\dfrac{1}{2x-3y}=\dfrac{1}{3}\Rightarrow2x-3y=3\end{matrix}\right.\)

giải hệ pt => \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{3}\end{matrix}\right.\)

Answer 2

Câu f)ĐKXĐ: 2x khác 3y hoặc 2x khác -3y

Đặt \(\dfrac{1}{2x+3y}\)=a \(\dfrac{1}{2x-3y}\)=b

hệ phương trình trở thành

\(\left\{{}\begin{matrix}2a-3b=-\dfrac{3}{5}\\9a+8b=\dfrac{67}{15}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}9a-\dfrac{27}{2}b=-\dfrac{27}{10}\\9a+8b=\dfrac{67}{15}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\dfrac{43}{2}b=\dfrac{43}{6}\\2a-3b=-\dfrac{3}{5}\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}b=\dfrac{1}{3}\\a=\dfrac{1}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x+3y=5\\2x-3y=3\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}4x=8\\2x+3y=5\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{3}\end{matrix}\right.\)

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